$A \cup B \equiv C $ then what is $A$ in terms of $B,C$?

I tried to use $A\cup B \equiv (A-B)\cup(A \cap B)\cup(B-A) $ to find a similar expression for $A \cap B$ but got nowhere.

From the elementary set theory that I did 40 or so years ago I don't recall any material on inverse of set theory operations i.e. union, intersection, complement, difference.

Complement is easy as it is it's own inverse $(A^C)^C=A$

not sure if inverse of difference operator is unique, for example, one can use $(A-B)\cup A \equiv A$ to construct one inverse of difference $ -X$.

when doing algebra, inverse operations are the first tricks to learn, but was the topic of inverse operations of set theory ever mentioned?

  • $\begingroup$ What does $\equiv$ mean in this context? Why aren't you saying they are equal? $\endgroup$ – fleablood Jan 20 '19 at 0:38

Inverse operators for union, intersection, or set difference are impossible in general.

Knowing $B$ and $C$ is not enough to determine what $A$ is. Consider for example $$ ? \cup \{1,2\} = \{1,2,3\} $$ Then either $\{1,3\}$ or $\{2,3\}$ would be possible solutions (and there are two more), so there's no operator that given just $\{1,2\}$ and $\{1,2,3\}$ can tell you "which of them $A$ really was".

On the other hand symmetric difference $$ A \mathbin{\triangle} B = (A\setminus B)\cup(B\setminus A) $$ has an inverse operation, namely itself: $(A\mathop\triangle B)\mathop\triangle B = A$.

If you consider the algebra of subsets of some universe $U$ under the operations $\triangle$ and $\cap$, you get a Boolean ring which satisfies the usual ring properties with $\triangle$ as addition and $\cap$ as multiplication. The ring's $0$ is $\varnothing$ and $1$ is $U$ itself.

This gives an opportunity to use more of the usual algebraic rules on sets. And you can express the remaining set operations in this vocabulary too: $$ A^\complement = 1 \mathop\triangle A \qquad\qquad A\cup B = A\mathop\triangle B \mathop\triangle(A\cap B) $$

What you lose by doing things this way is the nice duality between $\cup$ and $\cap$ and De Morgan's laws for sets.

(The multiplication still doesn't have an inverse, but it doesn't in general rings either, such as $\mathbb Z$).

  • $\begingroup$ Nice counter example! $\endgroup$ – fleablood Jan 20 '19 at 1:31
  • $\begingroup$ So if $C=A\triangle B$ then $C\triangle B=(A\triangle B)\setminus B \cup B\setminus(A\triangle B)=[(A\setminus B \cup B\setminus A)\setminus B ]\cup [B\setminus(A\setminus B \cup B\setminus A]=(A\setminus B)\cup B\setminus(B\setminus A) = (A\setminus B)\cup (B\cap A) = A$. .... nifty! $\endgroup$ – fleablood Jan 20 '19 at 2:04

It's not possible.

Given sets $C$ and $B$ there four states of being we can describe. $x $ in or not in $C$ and $x$ in or not in $B$ and we can only define sets by some combination of those conditions. Those conditions define the following $4$ disjoint basic sets and any set we can possibly be described in terms of $A$ and $B$ will be a union of these sets.

1) $\{x \not \in C, x\not \in B\} = C^c$ (as $B \subset C$)

2) $\{x \not \in C, x \in B\} = \emptyset$ (as $B \subset C$)

3) $\{x \in C, x \not \in B\} = C\setminus B$.

4) $\{x\in C, x \in B\} = C \cap B = B$ (as $B \subset C$).

Although $A$ is disjoint from 1) and 2) and 3) $\subset A$ set 4) will typically contain elements in $A$ and as well as elements not in $A$.

Hence in general we can not define $A$ solely on the conditions of whether they are or are not in $C$ or $B$.



We can list all possible sets possible to describe they are

The four above:

1) and [1 and 2] $C^c$, 2)$C^c\cap B = \emptyset$, 3) and [2 and 3] $C\setminus B$ and 4) and [2 and 4] $B$.

[1 and 3] and [1 and 2 and 3] = $C^c \cup C\setminus B = B^c$.

[1 and 4] and [1 and 2 and 4] = $C^c \cup B$

[3 and 4] and [2 and 3 and 4] = $(C\setminus B )\cup B = C$

[1 and 3 and 4] and [1 and 2 and 3 and 4] = Universal set.


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